Load Column is not governed by strength 2) Check base plate size a) Assume base plate size = Column width = 254 mm b) Contact pressure = Load/Area = 2463/0.254^"> Load Column is not governed by strength 2) Check base plate size a) Assume base plate size = Column width = 254 mm b) Contact pressure = Load/Area = 2463/0.254^">
C Assumed Fa 890 X 10 138.2
C Assumed Fa 890 X 10 138.2
C Assumed Fa 890 X 10 138.2
1.) A column that is 9 m. long is to carry a load of 890 kN. The member will be braced about both
principal axis at top and bottom and in addition will be braced about its minor axis at mid-height.
Using Fy = 345.5 MPa, design a section to carry the loads safely. Kx = Ky = 1.0
Properties of Steel Sections Available
Sections Area rx ry
W 8 x 40 7613 mm2 89.66 mm 51.82 mm
W 8 x 48 9097 mm2 91.69 mm 52.83 mm
W 10 x 49 9290 mm2 110.49 mm 64.52 mm
W 12 x 50 9484 mm2 131.57 mm 49.78 mm
Given:
L=9m
Fy = 345.5 MPa
Kx = Ky = 1.0 890 kN
C = 890 kN
y x
Solution:
a) Assume Fa (Max Fa = 0.6Fy)
Fa = 0.4 Fy
Fa = 0.4 (345.5)
Fa = 138.2 MPa
b) Gross – Area Required
C
Ag = Assumed Fa
890 x 103 4.5 m
Ag = 138.2
Ag = 6439.942 mm2
c) Try W 80 x 40
A = 7613 mm2
rx = 89.66 mm
ry = 51.82 mm
Slenderness Ratio:
KxL 1(9000)
rx
= 89.66 = 100.38
KyL 1(4500) 4.5 m
ry
= 51.82 = 86.64
𝐾𝐿
Use 𝑟 = 100.38 < 200 (SAFE)
2π2 (200000)
Cc = √ 345.5
Cc = 106.89 > 100.38 ; Therefore, Intermediate Column
Determine the Actual Allowable Fa:
2
(KL⁄r) Fy
Use: Fa = [1 - 2 Cc 2
] F.S.
3
5 3(KL⁄r) (KL⁄r)
F.S. = 3 + 8 Cc
- 8 Cc 3
5 3(100.38 ) (100.38 )3
F.S. = 3 + 8(106.89) - 8 (106.89)3
F.S. = 1.92
(100.38)2 345.5
Fa = [1 - ]
2 (106.89)2 1.92
Fa = 100.60 MPa
Determine the Capacity C
Cap. C = Fa Ag
Cap. C = 100.60 (7613)
Cap. C = 765866 N
Cap. C = 765.87 kN < 890 kN ; Therefore, try another section.
d) W 8 x 48
A = 9097 mm2
rx = 91.69 mm
ry = 52.83 mm
Slenderness Ratio:
KxL 1(9000)
= = 98.16
rx 91.69
KyL 1(4500)
= = 85.18
ry 52.83
𝐾𝐿
Use = 98.16 < 200 (SAFE)
𝑟
2π2 (200000)
Cc = √ 345.5
Cc = 106.89 > 98.16 ; Therefore, Intermediate Column
Determine the Actual Allowable Fa:
2
(KL⁄r) Fy
Use: Fa = [1 - ]
2 Cc 2 F.S.
3
5 3(KL⁄r) (KL⁄r)
F.S. = 3 + 8 Cc
- 8 Cc 3
5 3(98.16 ) (98.16)3
F.S. = 3 + 8(106.89) - 8 (106.89)3
F.S. = 1.91
(98.16)2 345.5
Fa = [1 - 2 (106.89)2
] 1.91
Fa = 104.62 MPa
Determine the Capacity C
Cap. C = Fa Ag
Cap. C = 104.62 (9097)
Cap. C = 951728 N
Cap. C = 951.73 kN < 890 kN ; SAFE
Therefore, use W 8 x 48
2.) Design the web members U1L1/U1’L1’ & U3M/U3’N of your riveted fink truss which are under
compression. Using two unequal leg angles long legs back to back straddling on a 12 mm gusset plate.
Use AISC specifications & ASTM A36 steel. K=1.0 (pin riveted connection.)
√15.62 +7.82
𝐿𝑜 𝑈1 = 4
= 4.36 m
7.8
𝜃 = tan−1 ( )
15.6
= 26.57°
Given:
C = 1600 kN
K = 1.0
Fy = 380 Mpa
Solution:
a) Determine Fa:
C
Fa = Ag L
1600(1000)
Fa = 21300
Fa = 75.12 MPa
b) Determine Cc:
2π2 E
Cc = √ Fy
2π2 (200000)
Cc = √
380
Cc = 101.93
c) Assume column is intermediate
KL
Let x =
R
x2 Fy
Fa =(1 - 2Cc 2 )( Fs )
5 3x x3
FS = 3 + 8Cc - 8Cc3
x2 Fy
Fa =(1 - 2Cc 2 )( 5 3x x 3
)
+ -
3 8Cc 8Cc 3
Solving for x:
𝑥2 380
75.12 = (1 − )( 𝑥3
)
2(101.93)2 5+ 3𝑥
−
3 8(101.93) 8(101.93)3
x1 = 1.686
x2 = - 1.775
x3 = 0.089
Solving for L using the largest value of x d) Assume column is long
KL 12π2 E
x= Fa= KL
R 23( )2
R
68.1(1.686)
L= 12π2 (200000)
1.0 75.12= 1.0L 2
23( )
L = 114.817 mm (unrealistic) 68.1
L= 7.974 m
4.) A W 250 x 73 is to serve as a pin-ended 12m long column. It is braced at mid-height with
respect to its weak axis.
Properties of W 250 x 73
A= 9280mm² 𝑟𝑥 = 110
d= 253mm 𝑟𝑦 = 64.7
𝑏𝑓 = 254mm 𝐼𝑥 = 113x106 𝑚𝑚4
𝑡𝑓 = 14.2mm 𝐼𝑦 = 38.8x106 𝑚𝑚4
a. Determine the slenderness ratio with respect to y-axis
b. Determine the Euler’s Buckling stress
c. Determine the allowable axial compressive load using a factor of safety of 2.5
Solution:
a) Slenderness ratio with respect to y-axis
𝐾𝐿 1(6000)
𝑟𝑦
= 64.7
𝑲𝑳
𝒓𝒚
= 92.74
Properties of W 12 x 79
B
d = 314.45 mm
bf = 306.83 mm
m
Given:
C = 2463.134 kN
fc’ = 28 MPa
Fy = 245 MPa
Solution: C
0.95 d
a) Size of Base Plate:
Allowable bearing stress of concrete:
Fp = 0.35 fc’
Fp = 0.35 (28)
Fp = 9.8 MPa
Trial area of base plate: m
C
A=
Fp
2463134 n
A = 9.8 0.80 bf n
A = 251340 mm2
Assume m = n Equivalent Base Plate
B = 2n + 0.80 bf Rectangular
B = 2n + 0.80(306.83) Section
B = 2n + 245.46 **
C = 2m + 0.95 d
C = 2m + 0.95(314.45)
C = 2m + 298.73 **
BC = 251340 mm2
(2m + 245.46) (2m + 298.73) = 251340
4m + 1088.38m + 73326.27 = 251340
4m2 + 1088.38m + 178013.73 = 0
m2 + 272.10m – 44503.43 = 0 b) Thickness of Base Plate:
m = 114.97 B = 2n + 245.46
480 = 2n + 245.46
Determine Size: n = 117.27
B = 2(114.97) + 245.46
B = 475.1 say 480 mm C = 2m + 298.73
530 = 2m + 298.73
C = 2(114.97) + 298.73
m = 115.64 mm
C = 528.67 say 530 mm
Use x = 117.27 (bigger value)
Use 480 mm x 530 mm Base Plate
3 Fp x 2
P t = √0.75 Fy
fp = BC
2463134 3 (9.8) (117.27)2
fp = t=√
480 (530) 0.75 (245)
L = √𝐴 = √489795 Equivalent
Rectangular Base Plate
L = 699.854 say 700mm
Section
700= 2n + 0.8(514)
n= 144.4mm
700= 2m + 0.95(437)
m= 142.425mm
3(7.35)(144.4)2
t= √ 0.75(248)
t= 49.718mm say 50mm
therefore use 700 x 700 x 50 mm base plate
7.) Determine the safe load of the column section shown, if it has a yield strength of 250 MPa. E =
200000 MPa. Use NSCP Specifications.
Solution:
Use Least I
70.3 x 106
r= √ 13379
r = 72.488 mm
L 6000
𝑆𝑅 = = = 82.77
r 72.488
5 3(SR) SR3
F.S. = 3 + - 8(C
3
8CC C)
5 3(82.77) (82.77)3
= + 8(125.66) − 8(125.66)3
3
F.S. =1.88
SR2 Fy
Fa =(1- 2 )( FS)
2CC
(82.77)2 250
= [1- 2(125.66)2] [ 1.88]
Fa = 104 MPa
P = FaA
P = 104(13379)(10-3)
P = 1391.4 kN
b) when L = 10 m:
L 10000
SR= r = 72.488
=138> CC ∴Long Column
12π2 E
Fa = 23(SR)2
12π2 (200000)
Fa = 23(138)2
Fa = 54.1 MPa
P = FaA = 54.1(13379)(10-3) = 723.8 kN
8.) A column is made of steel pipe with an outside diameter of 280 mm. The base plate of the column
rests on a circular base plate on concrete pedestal. The column is 3.6 m long and subjected to an axial
load of 900 kN. The allowable compressive stress in steel pipe is 65 MPa and the allowable bearing
stress in concrete pedestal is 12 MPa.
a. What is the required column thickness without exceeding its allowable compressive stress?
b. What is the required diameter of the steel base plate?
c. If the pipe is 10 mm thick, what is the effective slenderness ratio assuming that column is hinged
at both ends? (k=1.0)
Given:
D = 280 mm Fp = 12 MPa
L = 3.6 m Fa = 65 MPa
C = 900 kN
Solution:
C
a. Fa = A
𝜋(2802 − 𝑑2 ) 900(1000)
=
4 65
𝑑 = 246.517 𝑚𝑚
Since D – d = thickness t
therefore, Thickness = 33.483 mm 280 mm
C
b. Fp= d
Ap
KL 3600(1.0)
𝜋(𝐷 2 ) 900(1000) c. = √D2 +d2
R
= 4
4 12
KL 3600(1.0)
𝐝 = 𝟑𝟎𝟗. 𝟎𝟏𝟗 𝐦𝐦 =
R √2802 +246.5172
4
KL
R
= 38.60
Republic of the Philippines
Nueva Ecija University of Science and Technology
College of Engineering
Department of Civil Engineering
Cabanatuan City
Comprehensive Examination 3
Submitted by:
GROUP 2:
Lara Mariz B. Gatbonton
Justine Chris Lina
Kim Eucasion
Gabriel Lopez
BSCS 5-A
Submitted to: